Fast Fourier Transform

写在前面的..

感觉自己是应该学点新东西了.. 所以就挖个大坑，去学FFT了..

 

---

FFT是个啥？

坑已补上..

推荐去看黑书《算法导论》，讲的很详细

 

---

例题选讲

---

1.UOJ #34. 多项式乘法

这是FFT最裸的题目了 FFT就是拿来求这个东西的

没啥好讲的，把板子贴一下吧..

 

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <cmath>
using namespace std;
const int Maxn = 400010;
struct cp {
    double r, i;
    cp ( double r=0, double i=0 ) : r(r), i(i) {}
}A[Maxn], B[Maxn]; int an, bn, n;
cp operator + ( cp a, cp b ){ return cp ( a.r+b.r, a.i+b.i ); }
cp operator - ( cp a, cp b ){ return cp ( a.r-b.r, a.i-b.i ); }
cp operator * ( cp a, cp b ){ return cp ( a.r*b.r-a.i*b.i, a.r*b.i+a.i*b.r ); }
double pi = acos(-1);
void fft ( cp *a, int op ){
    int i, j, k;
    j = 0;
    for ( i = 0; i < n; i ++ ){
        if ( i < j ) swap ( a[i], a[j] );
        k = n>>1;
        while ( j & k ){ j -= k; k >>= 1; }
        j += k;
    }
    for ( i = 2; i <= n; i <<= 1 ){
        cp wn = cp ( cos (2.0*pi/i), op * sin (2.0*pi/i) );
        for ( j = 0; j < n; j += i ){
            cp w = cp ( 1, 0 );
            for ( k = j; k < j+i/2; k ++ ){
                cp x = a[k], y = w*a[k+i/2];
                a[k] = x+y; a[k+i/2] = x-y;
                w = w*wn;
            }
        }
    }
    if ( op == -1 ) for ( i = 0; i < n; i ++ ) a[i].r /= n;
}
int main (){
    int i, j, k;
    scanf ( "%d%d", &an, &bn );
    n = 1; while ( n < an+bn+1 ) n <<= 1;
    for ( i = 0; i <= an; i ++ ) scanf ( "%lf", &A[i].r );
    for ( i = 0; i <= bn; i ++ ) scanf ( "%lf", &B[i].r );
    fft ( A, 1 ); fft ( B, 1 );
    for ( i = 0; i < n; i ++ ) A[i] = A[i]*B[i];
    fft ( A, -1 );
    for ( i = 0; i <= an+bn; i ++ ) printf ( "%d ", (int)(A[i].r+0.5) );
    printf ( "\n" );
    return 0;
}

 

---

2.bzoj 3527: [Zjoi2014]力

这题在bzoj上没有题面，百度一下就行了，给个链接吧

化简一下公式：$$E_i=\sum\limits_{j<i}\dfrac{q_j}{(i-j)^2}-\sum\limits_{j>i}\dfrac{q_j}{(i-j)^2}$$

然后你设$g_i=\dfrac{1}{i^2}$，只考虑前面一部分的公式就可化成：$$E_i=\sum\limits_{j<i}q_j\times g_{i-j}$$

看是不是很眼熟，这就是很典型的FFT的模型

然后后面的就把整个数组反过来搞就是同样的道理..

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <cmath>
using namespace std;
const int Maxn = 100010;
struct cp {
    double r, i;
    cp ( double r=0.0, double i=0.0 ) : r(r), i(i) {}
}g[Maxn*4], h[Maxn*4];  int n, nn;
cp operator + ( cp a, cp b ){ return cp ( a.r+b.r, a.i+b.i ); }
cp operator - ( cp a, cp b ){ return cp ( a.r-b.r, a.i-b.i ); }
cp operator * ( cp a, cp b ){ return cp ( a.r*b.r-a.i*b.i, a.r*b.i+a.i*b.r ); }
double pi = acos (-1);
void fft ( cp *a, int op ){
    int i, j, k;
    for ( i = j = 0; i < n; i ++ ){
        if ( i < j ) swap ( a[i], a[j] );
        k = n >> 1;
        while ( j & k ) j -= k, k >>= 1;
        j += k;
    }
    for ( i = 2; i <= n; i <<= 1 ){
        cp wn = cp ( cos (2.0*pi/i), op * sin (2.0*pi/i) );
        for ( j = 0; j < n; j += i ){
            cp w = cp ( 1, 0 );
            for ( k = j; k < j+i/2; k ++ ){
                cp x = a[k], y = w*a[k+i/2];
                a[k] = x+y; a[k+i/2] = x-y;
                w = w*wn;
            }
        }
    }
    if ( op == -1 ) for ( i = 0; i < n; i ++ ) a[i].r /= n;
}
double q[Maxn], ans[Maxn];
int main (){
    int i, j, k;
    scanf ( "%d", &nn );
    for ( i = 1; i <= nn; i ++ ) scanf ( "%lf", &q[i] );
    n = 1; while ( n < 2*nn+1 ) n <<= 1;
    for ( i = 1; i <= nn; i ++ ){
        g[i].r = q[i];
        h[i].r = (double)1/(double(i)*i);
    }
    fft ( g, 1 ); fft ( h, 1 );
    for ( i = 0; i < n; i ++ ) g[i] = g[i]*h[i];
    fft ( g, -1 );
    for ( i = 1; i <= nn; i ++ ) ans[i] += g[i].r;
 
    for ( i = 0; i <= n; i ++ ) g[i].r = g[i].i = h[i].r = h[i].i = 0;
    for ( i = 1; i <= nn; i ++ ){
        g[i].r = q[nn-i+1];
        h[i].r = (double)1/(double(i)*i);
    }
    fft ( g, 1 ); fft ( h, 1 );
    for ( i = 0; i < n; i ++ ) g[i] = g[i]*h[i];
    fft ( g, -1 );
    for ( i = 1; i <= nn; i ++ ) ans[i] -= g[nn-i+1].r;
    for ( i = 1; i <= nn; i ++ ) printf ( "%lf\n", ans[i] );
    return 0;
}

---

3.bzoj 3160: 万径人踪灭

题意很长，但有用的不过那两句划线的..

仔细想想，如果$s_i==s_j$，那么就可以对$\dfrac{i+j}{2}$贡献答案了对吧

再想想，如果我已经知道了这$2n-1$（包括了中间的插缝）位置有$x$对数贡献了答案，在不考虑连续一段、取空集的情况下是可以有$2^x$种答案的

那么就在上面的基础上减去连续一段和空集的情况

连续一段的可以用manacher判断出，空集直接减去$1$就好了

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <cmath>
using namespace std;
const int Maxn = 100010;
const int Mod = 1e9+7;
struct cp {
    double r, i;
    cp ( double r=0.0, double i=0.0 ) : r(r), i(i) {}
}A[Maxn*4];
cp operator + ( cp a, cp b ){ return cp ( a.r+b.r, a.i+b.i ); }
cp operator - ( cp a, cp b ){ return cp ( a.r-b.r, a.i-b.i ); }
cp operator * ( cp a, cp b ){ return cp ( a.r*b.r-a.i*b.i, a.r*b.i+a.i*b.r ); }
int n, nn;
char s[Maxn*2]; int ans[Maxn*2];
double pi = acos (-1);
int _min ( int x, int y ){ return x < y ? x : y; }
void fft ( cp *a, int op ){
    int i, j, k;
    for ( i = j = 0; i < n; i ++ ){
        if ( i < j ) swap ( a[i], a[j] );
        k = n >> 1;
        while ( j & k ) j -= k, k >>= 1;
        j += k;
    }
    for ( i = 2; i <= n; i <<= 1 ){
        cp wn = cp ( cos (2.0*pi/i), op*sin(2.0*pi/i) );
        for ( j = 0; j < n; j += i ){
            cp w = cp ( 1, 0 );
            for ( k = j; k < j+i/2; k ++ ){
                cp x = a[k], y = w*a[k+i/2];
                a[k] = x+y; a[k+i/2] = x-y;
                w = w*wn;
            }
        }
    }
    if ( op == -1 ) for ( i = 0; i < n; i ++ ) a[i].r /= n;
}
int d[Maxn];
int rad[Maxn*2];
int main (){
    int i, j, k;
    scanf ( "%s", s );
    nn = strlen (s);
    d[0] = 1;
    for ( i = 1; i <= nn; i ++ ) d[i] = ( d[i-1] * 2 ) % Mod;
    n = 1; while ( n < 2*nn+1 ) n <<= 1;
    for ( i = 0; i < nn; i ++ ){
        if ( s[i] == 'a' ) A[i].r = 1;
    }
    fft ( A, 1 );
    for ( i = 0; i < n; i ++ ) A[i] = A[i]*A[i];
    fft ( A, -1 );
    for ( i = 0; i <= 2*(nn-1); i ++ ) ans[i] += ((int)(A[i].r+0.5)+1)/2;
 
    for ( i = 0; i < n; i ++ ) A[i].r = A[i].i = 0;
    for ( i = 0; i < nn; i ++ ){
        if ( s[i] == 'b' ) A[i].r = 1;
    }
    fft ( A, 1 );
    for ( i = 0; i < n; i ++ ) A[i] = A[i]*A[i];
    fft ( A, -1 );
    for ( i = 0; i <= 2*(nn-1); i ++ ) ans[i] += ((int)(A[i].r+0.5)+1)/2;
    int ret = 0;
    for ( i = nn-1; i >= 0; i -- ){
        s[i*2+2] = s[i];
        s[i*2+3] = '#';
    }
    s[0] = '$'; s[1] = '#'; s[2*nn+2] = '?';
    rad[0] = 0;
    k = 0; int mx = 0;
    for ( i = 1; i <= 2*nn; i ++ ){
        if ( mx > i ){ rad[i] = _min ( rad[2*k-i], mx-i ); }
        else rad[i] = 1;
        while ( s[i-rad[i]] == s[i+rad[i]] && i-rad[i] > 0 ) rad[i] ++;
        if ( i+rad[i]-1 > mx ){
            mx = i+rad[i]-1;
            k = i;
        }
    }
    for ( i = 0; i <= 2*(nn-1); i ++ ){
        ret = (ret+d[ans[i]]-rad[i+2]/2-1)%Mod;
    }
    printf ( "%d\n", ret );
    return 0;
}

---

4.bzoj 3509: [CodeChef] COUNTARI

来自la1la1la的题解：

1）先看一个比较sb的做法：枚举$j$，算出前面$i$的数的总数$sum[a[i]]$，然后找后面$k$，统计$sum[2*a[j]-a[k]]$

时间复杂度为$O(n^2)$

2）再看一个更sb的做法：枚举$j$，搞搞前面的和后面的，FFT计算，累加和为$2*a[j]$的方案

时间复杂度为$O(n^2logn)$

好，那么这题的正解就是把这两个sb做法合在一起

用一种分块的方法，假设我们把每一块分成每块大小为$S$的

枚举块内的$j$，用第一种方法暴力搞出块内的$i$、$k$的所有情况，时间复杂度为$O(nS)$

对于块外的使用第二种方法，即FFT来做，再枚举块内$j$累计答案，时间复杂度为$O(\frac{n}{S}mlogm)$，其中$m$是最大的数

总时间是$O(nS+\frac{n}{S}mlogm)$，这个时候锅就该甩给你们的高中老师了——均值不等式：$$a+b\geq 2\sqrt{ab}$$

那么这个时间就是大于等于$2n\sqrt{mlogm}$，什么时候能取到等于呢，继续甩锅

当$S=\sqrt{mlogm}$就是了，你把$m=30000$带进去估算一下（记住要带点常数进去..）

大概就是$S=1000$就差不多了 反正我就是这么打的..

 

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#define LL long long
using namespace std;
const int Maxn = 100010;
const int Maxm = 33000;
struct cp {
    double r, i;
    cp ( double r=0.0, double i=0.0 ) : r(r), i(i) {}
}A[Maxm*4], B[Maxm*4]; int An;
cp operator + ( cp a, cp b ){ return cp ( a.r+b.r, a.i+b.i ); }
cp operator - ( cp a, cp b ){ return cp ( a.r-b.r, a.i-b.i ); }
cp operator * ( cp a, cp b ){ return cp ( a.r*b.r-a.i*b.i, a.r*b.i+a.i*b.r ); }
int a[Maxn], n, S;
int _min ( int x, int y ){ return x < y ? x : y; }
int _max ( int x, int y ){ return x > y ? x : y; }
int ss[Maxm*2], ls[Maxm*2], rs[Maxm*2];
double pi = acos(-1);
void fft ( cp *a, int op ){
    int i, j, k;
    for ( i = j = 0; i < An; i ++ ){
        if ( i < j ) swap ( a[i], a[j] );
        k = An >> 1;
        while ( j & k ) j -= k, k >>= 1;
        j += k;
    }
    for ( i = 2; i <= An; i <<= 1 ){
        cp wn = cp ( cos (2.0*pi/i), op * sin (2.0*pi/i) );
        for ( j = 0; j < An; j += i ){
            cp w = cp ( 1, 0 );
            for ( k = j; k < j+i/2; k ++ ){
                cp x = a[k], y = w*a[k+i/2];
                a[k] = x+y; a[k+i/2] = x-y;
                w = w*wn;
            }
        }
    }
    if ( op == -1 ) for ( i = 0; i < An; i ++ ) a[i].r /= An;
}
int main (){
    int i, j, k;
    scanf ( "%d", &n );
    S = 1000;
    int mx = 0;
    for ( i = 1; i <= n; i ++ ){
        scanf ( "%d", &a[i] );
        rs[a[i]] ++;
        mx = _max ( mx, a[i] );
    }
    An = 1; while ( An < 2*mx+1 ) An <<= 1;
    LL ans = 0;
    for ( i = 1; i <= n; i += S ){
        int p = _min ( i+S-1, n );
        for ( j = i; j <= p; j ++ ) rs[a[j]] --;
        for ( j = i; j <= p; j ++ ){
            for ( k = j+1; k <= p; k ++ ){
                if ( 2*a[j]-a[k] > 0 ) ans += (LL)ss[2*a[j]-a[k]]+ls[2*a[j]-a[k]];
            }
            ss[a[j]] ++;
            for ( k = i; k <= j-1; k ++ ){
                if ( 2*a[j]-a[k] > 0 ) ans += (LL)rs[2*a[j]-a[k]];
            }
        }
        for ( j = 0; j < An; j ++ ){
            A[j].r = ls[j]; A[j].i = 0;
            B[j].r = rs[j]; B[j].i = 0;
        }
        fft ( A, 1 ); fft ( B, 1 );
        for ( j = 0; j < An; j ++ ) A[j] = A[j]*B[j];
        fft ( A, -1 );
        for ( j = i; j <= p; j ++ ){
            ans += (LL)(A[2*a[j]].r+0.5);
        }
        for ( j = i; j <= p; j ++ ) ss[a[j]] --, ls[a[j]] ++;
    }
    printf ( "%lld\n", ans );
    return 0;
}

 

---

5.hdu 5730 Shell Necklace

先把原题公式化简一下就是：$$dp_i=\sum\limits_{j=1}^{i-1}a_{i-j}\times dp_j$$

那么的话就是cdq分治搞一下再套一个FFT计算更新答案了..

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <cmath>
using namespace std;
const int Maxn = 100010;
const int Mod = 313;
struct cp {
    double r, i;
    cp ( double r=0.0, double i=0.0 ) : r(r), i(i) {}
}A[Maxn*4], B[Maxn*4]; int An;
cp operator + ( cp a, cp b ){ return cp ( a.r+b.r, a.i+b.i ); }
cp operator - ( cp a, cp b ){ return cp ( a.r-b.r, a.i-b.i ); }
cp operator * ( cp a, cp b ){ return cp ( a.r*b.r-a.i*b.i, a.r*b.i+a.i*b.r ); }
int a[Maxn], n, f[Maxn];
double pi = acos (-1);
void fft ( cp *a, int op ){
    int i, j, k;
    for ( i = j = 0; i < An; i ++ ){
        if ( i < j ) swap ( a[i], a[j] );
        k = An >> 1;
        while ( j & k ) j -= k, k >>= 1;
        j += k;
    }
    for ( i = 2; i <= An; i <<= 1 ){
        cp wn = cp ( cos (2*pi/i), op * sin (2*pi/i) );
        for ( j = 0; j < An; j += i ){
            cp w = cp ( 1, 0 );
            for ( k = j; k < j+i/2; k ++ ){
                cp x = a[k], y = w*a[k+i/2];
                a[k] = x+y; a[k+i/2] = x-y;
                w = w*wn;
            }
        }
    }
    if ( op == -1 ) for ( i = 0; i < An; i ++ ) a[i].r /= An;
}
void cdq ( int l, int r ){
    if ( l == r ) return;
    int mid = ( l + r ) >> 1;
    cdq ( l, mid );
    int i, j, k;
    int len = r-l;
    An = 1; while ( An < 2*len+1 ) An <<= 1;
    for ( i = l; i <= mid; i ++ ){
        A[i-l].r = f[i]; A[i-l].i = 0;
        B[i-l].r = a[i-l]; B[i-l].i = 0;
    }
    for ( i = mid+1; i <= r; i ++ ){
        A[i-l].r = 0; A[i-l].i = 0;
        B[i-l].r = a[i-l]; B[i-l].i = 0;
    }
    for ( i = len+1; i < An; i ++ ) A[i].r = A[i].i = B[i].r = B[i].i = 0;
    fft ( A, 1 ); fft ( B, 1 );
    for ( i = 0; i < An; i ++ ) A[i] = A[i]*B[i];
    fft ( A, -1 );
    for ( i = mid+1; i <= r; i ++ ){
        int ret = (int)(A[i-l].r+0.5);
        ret %= Mod;
        f[i] = ( f[i] + ret ) % Mod;
    }
    cdq ( mid+1, r );
}
int main (){
    int i, j, k;
    while ( scanf ( "%d", &n ) != EOF ){
        if ( n == 0 ) break;
        for ( i = 1; i <= n; i ++ ){
            scanf ( "%d", &a[i] ); a[i] %= Mod;
            f[i] = 0;
        }
        f[0] = 1;
        cdq ( 0, n );
        printf ( "%d\n", f[n] );
    }
    return 0;
}

---

6.hdu 5307 He is Flying

题意：有$n$段路，每段路长度为$s_i$，你从节点$i$到节点$j$，可以获得一个开心值$j−i+1$，然后问你，主人公走过了所有总长度为$s$的段，问你有多少开心值。

%%%la1la1la

累计一个前缀和$sum$，那么题意就变成走$sum_j-sum_i$的路程获得$j-i$的开心值

那么就把$j$和$-i$分开来算，卷积一下就好了..

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#define LL long long
#define LD long double
using namespace std;
const LL Maxn = 100010;
const LL Maxs = 50010;
struct cp {
    LD r, i;
    cp ( LD r=0.0, LD i=0.0 ) : r(r), i(i) {}
}A[Maxs*4], B[Maxs*4]; LL An;
cp operator + ( cp a, cp b ){ return cp ( a.r+b.r, a.i+b.i ); }
cp operator - ( cp a, cp b ){ return cp ( a.r-b.r, a.i-b.i ); }
cp operator * ( cp a, cp b ){ return cp ( a.r*b.r-a.i*b.i, a.r*b.i+a.i*b.r ); }
LL sum[Maxn], n, ans[Maxs];
const LD pi = acos (-1);
void fft ( cp *a, LL op ){
    LL i, j, k;
    for ( i = j = 0; i < An; i ++ ){
        if ( i < j ) swap ( a[i], a[j] );
        k = An >> 1;
        while ( j & k ) j -= k, k >>= 1;
        j += k;
    }
    for ( i = 2; i <= An; i <<= 1 ){
        for ( j = 0; j < An; j += i ){
            for ( k = j; k < j+i/2; k ++ ){
                cp x = a[k], y = cp ( cos (2*pi*(k-j)/i), op * sin (2*pi*(k-j)/i) ) * a[k+i/2];
                a[k] = x+y; a[k+i/2] = x-y;
            }
        }
    }
    if ( op == -1 ) for ( i = 0; i < An; i ++ ) a[i].r /= An;
}
LL ss[Maxn];
int main (){
    LL i, j, k, T;
    for ( i = 1; i <= 100000; i ++ ) ss[i] = ss[i-1]+i*(i+1)/2;
    scanf ( "%I64d", &T );
    while ( T -- ){
        scanf ( "%I64d", &n );
        LL lj = 0, ret = 0;
        for ( i = 1; i <= n; i ++ ){
            scanf ( "%I64d", &sum[i] );
            if ( sum[i] == 0 ) lj ++;
            else {
                ret += ss[lj];
                lj = 0;
            }
            sum[i] += sum[i-1];
        }
        ret += ss[lj];
        printf ( "%I64d\n", ret );
        for ( i = 0; i <= sum[n]; i ++ ) ans[i] = 0;
        An = 1; while ( An < sum[n]*2 ) An <<= 1;
        for ( i = 0; i < An; i ++ ) A[i].r = A[i].i = B[i].r = B[i].i = 0;
        for ( i = 0; i <= n; i ++ ){
            A[sum[n]-sum[i]].r += i;
            B[sum[i]].r += 1;
        }
        fft ( A, 1 ); fft ( B, 1 );
        for ( i = 0; i < An; i ++ ) A[i] = A[i]*B[i];
        fft ( A, -1 );
        for ( i = 0; i <= sum[n]; i ++ ) ans[i] = -(LL)(A[sum[n]+i].r+0.5);
        for ( i = 0; i < An; i ++ ) A[i].r = A[i].i = B[i].r = B[i].i = 0;
        for ( i = 0; i <= n; i ++ ){
            A[sum[n]-sum[i]].r += 1;
            B[sum[i]].r += i;
        }
        fft ( A, 1 ); fft ( B, 1 );
        for ( i = 0; i < An; i ++ ) A[i] = A[i]*B[i];
        fft ( A, -1 );
        for ( i = 0; i <= sum[n]; i ++ ) ans[i] += (LL)(A[sum[n]+i].r+0.5);
        for ( i = 1; i <= sum[n]; i ++ ) printf ( "%I64d\n", ans[i] );
    }
    return 0;
}

---

7.hdu 4609/bzoj 3513 3-idiots

题意：给你$n$条边，问你任取三条边能组成三角形的概率

按权值卷积后，把边排序

对于任意一条边你都能知道其余两条边加起来大于它的方案数

考虑第$i$条边是最长边，那么就要减去一条选大的一条选小的、两条都选大的方案数，这些都可以用组合公式算得..

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#define LL long long
using namespace std;
const LL Maxn = 100010;
struct cp {
    double r, i;
    cp ( double r=0.0, double i=0.0 ) : r(r), i(i) {}
}A[Maxn*4]; LL An;
cp operator + ( cp a, cp b ){ return cp ( a.r+b.r, a.i+b.i ); }
cp operator - ( cp a, cp b ){ return cp ( a.r-b.r, a.i-b.i ); }
cp operator * ( cp a, cp b ){ return cp ( a.r*b.r-a.i*b.i, a.r*b.i+a.i*b.r ); }
LL a[Maxn], n;
LL ans[Maxn*2];
const double pi = acos (-1);
LL _max ( LL x, LL y ){ return x > y ? x : y; }
void fft ( cp *a, LL op ){
    LL i, j, k;
    for ( i = j = 0; i < An; i ++ ){
        if ( i < j ) swap ( a[i], a[j] );
        k = An >> 1;
        while ( j & k ) j -= k, k >>= 1;
        j += k;
    }
    for ( i = 2; i <= An; i <<= 1 ){
        cp wn = cp ( cos (2*pi/i), op * sin (2*pi/i) );
        for ( j = 0; j < An; j += i ){
            cp w = cp ( 1, 0 );
            for ( k = j; k < j+i/2; k ++ ){
                cp x = a[k], y = w*a[k+i/2];
                a[k] = x+y; a[k+i/2] = x-y;
                w = w*wn;
            }
        }
    }
    if ( op == -1 ) for ( i = 0; i < An; i ++ ) A[i].r /= An;
}
int main (){
    LL i, j, k, T;
    scanf ( "%I64d", &T );
    while ( T -- ){
        scanf ( "%I64d", &n );
        LL mx = 0;
        for ( i = 1; i <= n; i ++ ){
            scanf ( "%I64d", &a[i] );
            mx = _max ( mx, a[i] );
        }
        An = 1;    while ( An < 2*mx+1 ) An <<= 1;
        for ( i = 0; i < An; i ++ ) A[i].r = A[i].i = 0;
        for ( i = 1; i <= n; i ++ ){
            A[a[i]].r += 1;
        }
        fft ( A, 1 );
        for ( i = 0; i < An; i ++ ) A[i] = A[i]*A[i];
        fft ( A, -1 );
        for ( i = 1; i <= mx*2; i ++ ) ans[i] = (LL)(A[i].r+0.5);
        for ( i = 1; i <= n; i ++ ) ans[2*a[i]] --;
        for ( i = 1; i <= mx*2; i ++ ) ans[i] /= 2;
        for ( i = mx*2-1; i >= 1; i -- ) ans[i] += ans[i+1];
        sort ( a+1, a+n+1 );
        LL ret = 0;
        for ( i = 1; i <= n; i ++ ){
            ret += (LL)ans[a[i]+1]-n+1;
            ret -= (LL)(n-i)*(i-1);
            ret -= (LL)(n-i)*(n-i-1)/2;
        }
        LL zs = n*(n-1)*(n-2)/6;
        printf ( "%.7lf\n", (double)ret/zs );
    }
    return 0;
}

---

8.hdu 5751 Eades

题意：

 

Peter有一个序列$a_1, a_2, ..., a_n$. 定义$g(l,r)$表示子序列$a_{l},a_{l+1},...,a_{r}$的最大值, $f(l,r)=\displaystyle\sum_{i=l}^{r}[a_i = g(l,r)]$. 注意$[\text{condition}] = 1$当且仅当$\text{condition}$是true, 否则$[\text{condition}] = 0$.

 

对于每个整数$k \in \{1, 2, ..., n\}$, Peter想要知道有多少整数对$l$和$r$ $(l \le r)$满足$f(l,r)=k$.

 

官方题解：

 

枚举每个数$x$, 考虑这个数成为最大值的对每个$z(\cdot)$的贡献. 对于每个数$x$, 你都可以求出若干个极大区间$[l_x,r_x]$, 表示这个$x$在区间内是最大值, 不妨假设这个$x$在$[l_x,r_x]$出现了$m$次, 每个位置分别为$p_1,p_2,...,p_m$, 那么就可以转化成一个长度为$m+1$的序列$c_0,c_2,...,c_m$. 其中$c_0=p_1-l+1$ $c_{m}=r-p_m+1$ $c_i=p_{i+1}-p_{i}, 1 \leq i < m$

 

于是对$z_k$的贡献就是$z_k=\sum_{i=0}^{m}c_i \cdot c_{i+k}$

 

这个其实就是$c_0,c_1,...,c_m$和$c_m,c_{m-1},...,c_0$的卷积, 做一次fft就好了.

时间复杂度分析：因为每一个数只有可能进入一次做fft，所以总时间均摊$O(nlogn)$

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <cmath>
#define LL long long
using namespace std;
const LL Maxn = 100010;
struct cp {
    double r, i;
    cp ( double r=0.0, double i=0.0 ) : r(r), i(i) {}
}A[Maxn*4], B[Maxn*4]; LL An;
cp operator + ( cp a, cp b ){ return cp ( a.r+b.r, a.i+b.i ); }
cp operator - ( cp a, cp b ){ return cp ( a.r-b.r, a.i-b.i ); }
cp operator * ( cp a, cp b ){ return cp ( a.r*b.r-a.i*b.i, a.r*b.i+a.i*b.r ); }
LL a[Maxn], maxx[Maxn*4], n;
vector <LL> vec[Maxn];
LL _max ( LL x, LL y ){ return x > y ? x : y; }
void bulid_tree ( LL now, LL L, LL R ){
    if ( L < R ){
        LL mid = ( L + R ) >> 1, lc = now*2, rc = now*2+1;
        bulid_tree ( lc, L, mid );
        bulid_tree ( rc, mid+1, R );
        maxx[now] = _max ( maxx[lc], maxx[rc] );
    }
    else maxx[now] = a[L];
}
LL query ( LL now, LL L, LL R, LL l, LL r ){
    if ( L == l && R == r ) return maxx[now];
    LL mid = ( L + R ) >> 1, lc = now*2, rc = now*2+1;
    if ( r <= mid ) return query ( lc, L, mid, l, r );
    else if ( l > mid ) return query ( rc, mid+1, R, l, r );
    else return _max ( query ( lc, L, mid, l, mid ), query ( rc, mid+1, R, mid+1, r ) );
}
LL ans[Maxn];
const double pi = acos (-1);
void fft ( cp *a, LL op ){
    LL i, j, k;
    for ( i = j = 0; i < An; i ++ ){
        if ( i < j ) swap ( a[i], a[j] );
        k = An >> 1;
        while ( j & k ) j -= k, k >>= 1;
        j += k;
    }
    for ( i = 2; i <= An; i <<= 1 ){
        cp wn = cp ( cos (2*pi/i), op * sin (2*pi/i) );
        for ( j = 0; j < An; j += i ){
            cp w = cp ( 1, 0 );
            for ( k = j; k < j+i/2; k ++ ){
                cp x = a[k], y = w*a[k+i/2];
                a[k] = x+y; a[k+i/2] = x-y;
                w = w*wn;
            }
        }
    }
    if ( op == -1 ) for ( i = 0; i < An; i ++ ) a[i].r /= An;
}
void solve ( LL l, LL r ){
    if ( l > r ) return;
    LL i, j, k;
    LL Max = query ( 1, 1, n, l, r );
    LL x = lower_bound ( vec[Max].begin (), vec[Max].end (), l ) - vec[Max].begin ();
    LL y = upper_bound ( vec[Max].begin (), vec[Max].end (), r ) - vec[Max].begin ();
    An = 1; while ( An < 2*(y-x)+1 ) An <<= 1;
    for ( i = 0; i < An; i ++ ) A[i].r = A[i].i = B[i].r = B[i].i = 0;
    A[0].r = vec[Max][x]-l+1;
    for ( i = x; i < y-1; i ++ ){
        A[i-x+1].r = vec[Max][i+1]-vec[Max][i];
    }
    A[y-x].r = r-vec[Max][y-1]+1;
    for ( i = 0; i <= y-x; i ++ ) B[i].r = A[y-x-i].r;
    fft ( A, 1 ); fft ( B, 1 );
    for ( i = 0; i < An; i ++ ) A[i] = A[i]*B[i];
    fft ( A, -1 );
    for ( i = 1; i <= y-x; i ++ ) ans[i] += (LL)(A[i+y-x].r+0.5);
    solve ( l, vec[Max][x]-1 );
    for ( i = x; i < y-1; i ++ ) solve ( vec[Max][i]+1, vec[Max][i+1]-1 );
    solve ( vec[Max][y-1]+1, r );
}
int main (){
    LL i, j, k, T;
    scanf ( "%I64d", &T );
    while ( T -- ){
        scanf ( "%I64d", &n );
        for ( i = 1; i <= n; i ++ ) vec[i].clear ();
        for ( i = 1; i <= n; i ++ ){ scanf ( "%I64d", &a[i] ); vec[a[i]].push_back (i); }
        bulid_tree ( 1, 1, n );
        for ( i = 1; i <= n; i ++ ) ans[i] = 0;
        solve ( 1, n );
        LL ret = 0;
        for ( i = 1; i <= n; i ++ ){
            ret += ans[i]^i;
        }
        printf ( "%I64d\n", ret );
    }
    return 0;
}

---

 9.bzoj 4332: JSOI2012 分零食

简化题意：把$m$分到$n$个位置，允许后缀为$0$，每个位置的值为$f(i)=Oi^2+Si+U$，其中$i$是该位置分到的数，问你所有情况的权值积的和

找到三种做法：

YJQ大爷

WerKeyTom （讲道理不是很理解..）

某大神 （比较厉害的做法..）

我就是用YJQ大爷的方法的.. 他的题解讲的不是很详细就来补充一下吧

设$f_{i,j}$表示把$j$分到$i$个位置而且全部填满没有后缀$0$的答案

$g_{i,j}$表示把$j$分到$i$个位置而且一定至少有一个空没填的答案

由于$n\leq 10^8$，所以肯定是要用倍增的方法

然后就是可以这么推：$$f_{i,j}=\sum\limits_{k=1}^{j-1}f_{\frac{i}{2},j-k}\cdot f_{\frac{i}{2},k}$$

$$g_{i,j}=\sum\limits_{k=1}^{j-1}g_{\frac{i}{2},j-k}\cdot f_{\frac{i}{2},k}$$

大概就是这样..

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <cmath>
using namespace std;
const int Maxn = 10010;
int m, P, n, O, S, U;
struct cp {
    double r, i;
    cp ( double r=0.0, double i=0.0 ) : r(r), i(i) {}
}A[Maxn*4], B[Maxn*4];
cp operator + ( cp a, cp b ){ return cp ( a.r+b.r, a.i+b.i ); }
cp operator - ( cp a, cp b ){ return cp ( a.r-b.r, a.i-b.i ); }
cp operator * ( cp a, cp b ){ return cp ( a.r*b.r-a.i*b.i, a.r*b.i+a.i*b.r ); }
int f[Maxn*4], g[Maxn*4], ansf[Maxn*4], ansg[Maxn*4], An;
int rev[Maxn*4];
const double pi = acos (-1);
void fft ( cp *a, int op ){
    int i, j, k;
    for ( i = 0; i < An; i ++ ) if ( i < rev[i] ) swap ( a[i], a[rev[i]] );
    for ( i = 2; i <= An; i <<= 1 ){
        cp wn = cp ( cos (2*pi/i), op * sin (2*pi/i) );
        for ( j = 0; j < An; j += i ){
            cp w = cp ( 1, 0 );
            for ( k = j; k < j+i/2; k ++ ){
                cp x = a[k], y = w*a[k+i/2];
                a[k] = x+y; a[k+i/2] = x-y;
                w = w*wn;
            }
        }
    }
    if ( op == -1 ) for ( i = 0; i < An; i ++ ) a[i].r /= An;
}
int main (){
    int i, j, k;
    scanf ( "%d%d%d%d%d%d", &m, &P, &n, &O, &S, &U );
    An = 1; while ( An < 2*m+1 ) An <<= 1;
    j = 0;
    for ( i = 0; i < An; i ++ ){
        rev[i] = j;
        k = An >> 1;
        while ( j & k ) j -= k, k >>= 1;
        j += k;
    }
    g[0] = 1;
    for ( i = 1; i <= m; i ++ ){
        f[i] = (((O*i)%P*i)%P+(S*i)%P+U)%P;
    }
    ansf[0] = 1; ansg[0] = 0;
    while (n){
        if ( n & 1 ){
            for ( i = 0; i <= m; i ++ ) A[i].r = ansf[i], B[i].r = g[i], A[i].i = B[i].i = 0;
            for ( i = m+1; i < An; i ++ ) A[i].r = B[i].r = A[i].i = B[i].i = 0;
            fft ( A, 1 ); fft ( B, 1 );
            for ( i = 0; i < An; i ++ ) A[i] = A[i]*B[i];
            fft ( A, -1 );
            for ( i = 0; i <= m; i ++ ) ansg[i] = ( ansg[i] + ((int)(A[i].r+0.5))%P ) % P;
            
            for ( i = 0; i <= m; i ++ ) A[i].r = ansf[i], B[i].r = f[i], A[i].i = B[i].i = 0;
            for ( i = m+1; i < An; i ++ ) A[i].r = B[i].r = A[i].i = B[i].i = 0;
            fft ( A, 1 ); fft ( B, 1 );
            for ( i = 0; i < An; i ++ ) A[i] = A[i]*B[i];
            fft ( A, -1 );
            for ( i = 0; i <= m; i ++ ) ansf[i] = ((int)(A[i].r+0.5))%P;
        }
        for ( i = 0; i <= m; i ++ ) A[i].r = f[i], B[i].r = g[i], A[i].i = B[i].i = 0;
        for ( i = m+1; i < An; i ++ ) A[i].r = B[i].r = A[i].i = B[i].i = 0;
        fft ( A, 1 ); fft ( B, 1 );
        for ( i = 0; i < An; i ++ ) A[i] = A[i]*B[i];
        fft ( A, -1 );
        for ( i = 0; i <= m; i ++ ) g[i] = ( g[i] + ((int)(A[i].r+0.5))%P ) % P;
        
        for ( i = 0; i <= m; i ++ ) A[i].r = f[i], A[i].i = 0;
        for ( i = m+1; i < An; i ++ ) A[i].r = A[i].i = 0;
        fft ( A, 1 );
        for ( i = 0; i < An; i ++ ) A[i] = A[i]*A[i];
        fft ( A, -1 );
        for ( i = 0; i <= m; i ++ ) f[i] = ((int)(A[i].r+0.5))%P;
        n >>= 1;
    }
    printf ( "%d\n", (ansf[m]+ansg[m])%P );
    return 0;
}

---

 10.bzoj 3456: 城市规划

这是一道比较厉害的NTT题目.. 我不是很会解释，建议去看看WerKeyTom的blog

这道题的时间也比较坑，我卡了很久的常数才过..

 

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <ctime>
#include <iostream>
#define LL long long
using namespace std;
const LL Maxn = 130010;
const LL Mod = 1004535809;
const LL G = 3;
LL A[Maxn*4], B[Maxn*4], An;
LL pow ( LL x, LL k ){
    LL ret = 1;
    while (k){
        if ( k & 1 ) ret = (ret*x)%Mod;
        x = (x*x)%Mod;
        k >>= 1;
    }
    return ret;
}
LL v[Maxn], f[Maxn], jc[Maxn], fjc[Maxn], n, inv[Maxn], cf[Maxn*4], fcf[Maxn*4], p[Maxn];
void dft ( LL *a, LL op ){
    LL i, j, k;
    for ( i = j = 0; i < An; i ++ ){
        if ( i < j ) swap ( a[i], a[j] );
        k = An >> 1;
        while ( j & k ) j -= k, k >>= 1;
        j += k;
    }
    for ( i = 2; i <= An; i <<= 1 ){
        LL wn = op > 0 ? cf[i] : fcf[i];
        LL w = 1;
        for ( k = 0; k < i/2; k ++ ){
            for ( j = 0; j < An; j += i ){
                LL x = a[k+j], y = (w*a[k+j+i/2])%Mod;
                a[k+j] = (x+y)%Mod; a[k+j+i/2] = (x-y+Mod)%Mod;
            }
            w = (w*wn)%Mod;
        }
    }
    if ( op == -1 ) for ( i = 0; i < An; i ++ ) a[i] = (a[i]*inv[An])%Mod;
}
void cdq ( LL l, LL r ){
    if ( l == r ){ f[l] = (v[l]-(jc[l-1]*f[l])%Mod+Mod)%Mod; return; }
    LL mid = ( l + r ) >> 1, i;
    cdq ( l, mid );
    LL len = r-l+1;
    An = 1; while ( An < 2*len+1 ) An <<= 1;
    for ( i = 0; i < An; i ++ ) A[i] = B[i] = 0;
    for ( i = 0; i < len; i ++ ) B[i] = p[i];
    for ( i = 0; i < mid-l+1; i ++ ) A[i] = (f[i+l]*fjc[i+l-1])%Mod; 
    dft ( A, 1 ); dft ( B, 1 );
    for ( i = 0; i < An; i ++ ) A[i] = (A[i]*B[i])%Mod;
    dft ( A, -1 );
    for ( i = mid+1; i <= r; i ++ ) f[i] = (f[i]+A[i-l])%Mod;
    cdq ( mid+1, r );
}
int c[Maxn];
int main (){
    LL i, j, k;
    scanf ( "%lld", &n );
    jc[0] = 1; fjc[0] = 1; v[0] = 1;
    c[0] = 1; for ( i = 1; i <= n; i ++ ) c[i] = (c[i-1]*2)%Mod;
    for ( i = 1; i <= n; i ++ ){
        v[i] = ( v[i-1] * c[i-1] ) % Mod;
        jc[i] = (jc[i-1]*i)%Mod;
    }
    fjc[n] = pow ( jc[n], Mod-2 );
    for ( i = n-1; i >= 1; i -- ){
        fjc[i] = (fjc[i+1]*(i+1))%Mod;
        p[i] = (v[i]*fjc[i])%Mod;
    }
    An = 1; while ( An < 2*n+1 ) An <<= 1;
    for ( i = 2; i <= An; i <<= 1 ) cf[i] = pow ( G, (Mod-1)/i ), fcf[i] = pow ( G, Mod-1-(Mod-1)/i ), inv[i] = pow ( i, Mod-2 );
    cdq ( 1, n );
    printf ( "%lld\n", f[n] );
    return 0;
}

 

---

11.bzoj 4555: [Tjoi2016&Heoi2016]求和

 

把公式化一下基本就出来了，这里给出几个公式：

斯特林数化简公式：$$S(n,m)=\dfrac{1}{m!}\sum\limits_{k=0}^{m}(-1)^k\cdot C_m^k\cdot (m-k)^n$$

等比数列求和公式：$$Sum=\dfrac{a(q^k-1)}{q-1}$$

那么最后化成的公式为：$$f(n)=\sum\limits_{j=0}^n2^j\cdot j!\sum\limits_{k=0}^jg(k)\cdot h(j-k)$$

$$g(i)=\dfrac{(-1)^i}{i!},\ h(i)=\dfrac{\sum_{k=0}^{n}i^k}{i!}$$

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#define LL long long
using namespace std;
const LL Maxn = 100010;
const LL Mod = 998244353;
const LL G = 3;
LL jc[Maxn], inv[Maxn], invn;
LL pow ( LL x, LL k ){
    LL ret = 1;
    while (k){
        if ( k & 1 ) ret = (ret*x)%Mod;
        x = (x*x)%Mod;
        k >>= 1;
    }
    return ret;
}
LL A[Maxn*4], B[Maxn*4], An;
LL n;
LL dft ( LL *a, LL op ){
    LL i, j, k;
    for ( i = j = 0; i < An; i ++ ){
        if ( i < j ) swap ( a[i], a[j] );
        k = An >> 1;
        while ( j & k ) j -= k, k >>= 1;
        j += k;
    }
    for ( i = 2; i <= An; i <<= 1 ){
        LL wn = pow ( G, op > 0 ? (Mod-1)/i : Mod-1-(Mod-1)/i );
        for ( j = 0; j < An; j += i ){
            LL w = 1;
            for ( k = j; k < j+i/2; k ++ ){
                LL x = a[k], y = (w*a[k+i/2])%Mod;
                a[k] = (x+y)%Mod; a[k+i/2] = (x-y+Mod)%Mod;
                w = (w*wn)%Mod;
            }
        }
    }
    if ( op == -1 ) for ( i = 0; i < An; i ++ ) a[i] = (a[i]*invn)%Mod;
}
int main (){
    LL i, j, k;
    scanf ( "%lld", &n );
    An = 1; while ( An < 2*n+1 ) An <<= 1;
    jc[0] = jc[1] = 1;
    for ( i = 2; i <= n; i ++ ) jc[i] = (jc[i-1]*i)%Mod;
    inv[n] = pow ( jc[n], Mod-2 );
    for ( i = n-1; i >= 1; i -- ) inv[i] = (inv[i+1]*(i+1))%Mod;
    A[0] = 1;
    for ( i = 1; i <= n; i ++ ) A[i] = i % 2 == 1 ? Mod-inv[i] : inv[i];
    B[0] = 1; B[1] = n+1;
    for ( i = 2; i <= n; i ++ ) B[i] = ( ( ( pow ( i, n+1 ) - 1 ) * pow ( i-1, Mod-2 ) ) % Mod * inv[i] ) % Mod;
    invn = pow ( An, Mod-2 );
    dft ( A, 1 ); dft ( B, 1 );
    for ( i = 0; i < An; i ++ ) A[i] = (A[i]*B[i])%Mod;
    dft ( A, -1 );
    LL s = 1, ans = 0;
    for ( i = 0; i <= n; i ++ ){
        ans = ( ans + ((s*jc[i])%Mod*A[i])%Mod ) % Mod;
        s = (s*2)%Mod;
    }
    printf ( "%lld\n", ans );
    return 0;
}

 

---

参考资料

picks的blog